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Unformatted text preview: with n1 degrees of freedom. So, for the desired probability, we can write. Pr(0 . 03 < S < . 07) = Pr(0 . 0009 < S 2 < . 0049) = Pr . 0009 19 . 0025 < X 2 < . 0049 19 . 0025 = Pr(6 . 84 < X 2 < 37 . 24) = 0 . 995. 007 = 0 . 988 . 4. We know that P N p, q p (1p ) n . So for the probability in question, we obtain Pr( P < . 60) = . 60. 70 q . 70 . 30 200 = (3 . 09) = 0 . 001 . 1...
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 Spring '08
 Perevalov

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